Answer:
C- the breakdown of rock into sediment
Explanation:
Because sedimentary rocks have layers and are made of of those layers which are made up of the breakdown of rock into sediment.
hope this helps:)
The number of moles of silver oxide (I) needed to produce 4 moles of silver is 2 moles
<h3>Stoichiometry </h3>
From the question, we are to determine the number of moles of silver oxide (I) needed to produce 4 moles of silver
First, we will write the balaced chemical equation for the decomposition of silver oxide (I)
2Ag₂O(s) → 4Ag(s) + O₂(g)
This means, 2 moles of silver oxide (I) [Ag₂O] decomposes to give 4 moles of <u>silver </u>and 1 mole of oxygen gas.
From the <em>balanced chemical equation</em>, it is easy to deduce the number of moles of silver oxide (I) that would give 4 moles of silver.
Hence, the number of moles of silver oxide (I) needed to produce 4 moles of silver is 2 moles
Learn more on Stoichiometry here: brainly.com/question/18834543
Answer: well 18°C is really cool the mass of the water sample would be 28242 yeah that's right
Explanation:
Answer:
True
Explanation:
A metal , has 1 electron on the last level, so it can become (A^+1)
C, non-metal, has 7 electrons on the last level, so it can become (C^-1)
Compound AC is possible
We are given the mass spectrum data for this compound which has a molecular ion peak of m+ = 91.043 m/z. When we have an m+ peak that is an odd number, that suggests that there are an odd number of nitrogens, in this case we'll assume 1 nitrogen atom to start. Nitrogen has a mass of 14 so we will substract that from our initial value.
91- 14 (1N) = 77 m/z
We are also told that there are carbon, hydrogens and oxygens present, so we will assume there is at least one oxygen which has a mass of 16 and subtract that value.
77 - 16 (1 O) = 61 m/z
Now we will try to get as close as possible to the remaining mass with carbons that has a mass of 12, and fill the remaining mass with hydrogens that have a mass of 1.
61 / 12 = 5
5 x 12 = 60
61 - 60 (5 C) = 1 m/z and this leaves us with 1 H.
The current formula would be C₅HON, but this structure is impossible since we do not have enough hydrogens to satisfy the carbons. So we can try to use 4 carbons instead and fill the rest with hydrogens.
4 x 12 = 48
61 - 48 (4 C) = 13 m/z and this leaves us with 13 H.
The current formula would be C₄H₁₃ON. The most hydrogens we can have in a compound is 2n+2 where n is the number of carbons. So with 4 carbons the most hydrogens we could have is 10. Therefore, our formula has too many hydrogens and also cannot work. So we cannot make up the remaining mass with carbons and hydrogens, therefore, we should add another oxygen before working with carbons and hydrogens.
61 - 16 (1 O) = 45 m/z
45/ 12 = 3.75
3 x 12 = 36
45 - 36 (3 C) = 9 m/z which gives us 9 hydrogens left.
The current formula is now C₃H₉O₂N. To test if this formula works we can calculate the double bond equivalents (DBE), also known as degrees of unsaturation.
DBE = C - H/2 + N/2 + 1 = 3 - (9/2) + (1/2) + 1 = 0
A value of 0 DBE tells us that there are no double bonds in this molecule but that the formula is a possibility so:
C₃H₉O₂N = 91 m/z
